> It is an old theorem that in Hex, once the board has been completely> filled in with two colours, there *must* be a winning path for one> or other of them.> > Now, I can prove this easily enough mathematically, but I'm wondering> if there is a simple proof, or proof outline, that would be> understandable and reasonably convincing to the intelligent layman.> > Can anyone help out please?>

Here's what I'm thinking...

Suppose you have red going top-bottom and blue going left-right. If red does not win, then there must be a path that divides the board into to pieces, a top and a bottom piece. Red cannot be in any position in that path, so it must be blue. Thus blue wins. Rotate pi/2 and switch colors. QED